6

I have a device that requires 1.21 Gigawatts over one second to fully activate. Upon activation the device is disconnected automatically. The device can accept the power at any voltage, and will simply present itself as a load that will consume the required power in one second regardless of the source voltage.

I have an energy source, essentially modeled as a capacitor, with an effective capacitance of 159µF.

Assuming I charge this energy source to 7.69 x 10^15 Volts, and there are negligible losses between the power source and device, how many device activations can I achieve before I must recharge it?

Community
  • 1
Adam Davis
  • 20,339
  • 7
  • 59
  • 95
  • 5 * 10^17 times :) – HL-SDK May 23 '14 at 19:31
  • 1
    (1/2 CV^2 / (1.2*10^9 watts * 1 second)) C = 159uF, V = 7.7PV math by the seat of my pants – HL-SDK May 23 '14 at 19:32
  • 3
    What the h*ll is a jigga-watt?! :) [youtube](http://www.youtube.com/watch?v=I5cYgRnfFDA) – bitsmack May 23 '14 at 19:51
  • 2
    Now, that's a real world question, finally. – Dzarda May 23 '14 at 19:57
  • 2
    @HL-SDK: I think you slipped a decimal place somwhere. I get 3.88e18 times. I just wanna see him do it with "negligible losses"! :-) – Dave Tweed May 23 '14 at 19:58
  • This isn't the second grade anymore, so please stop writing all those zeros. It's annoying to have to count zeros to understand a number. Also, where are you going to get 7.7e15 volts from, or keep it from arcing even if you had it? At 1kV/mm air breakdown, even opposite points on the surface of the earth aren't anywhere near far enough apart. – Olin Lathrop May 23 '14 at 20:28
  • 2
    Well, to be fair, the picture linked in the OP does show some arcing. – WhatRoughBeast May 23 '14 at 20:55
  • 1.21 gigawatts? Is that you Doc Brown? – Passerby May 23 '14 at 23:04
  • typical grumpy Olin. Perhaps we should do 7.69e15 - 1 volts so space doesn't break down. I'll check my math tomorrow and post an answer. – HL-SDK May 24 '14 at 00:41

2 Answers2

4

AT first I thought would be a bogus question as the energy source might not be available (in the right position). So I asked a friend

enter image description here to take a picture.

enter image description here

With that data I visited the NASA website and determined that your energy source was waxing gibbous. The next full moon would have been Oct 28 so slightly off of peak.

http://eclipse.gsfc.nasa.gov/phase/phases1901.html

enter image description here

If you need a repeatable experiment I would suggest an alternative energy source. enter image description here

I've never actually used this product, but I suspect that it is a cheap alibaba knock off . I don't see how to get such a small device without the appropriate shielding and slow neutron absorption (for protection). Please be careful.

I know I haven't answered your question "how many actuations" but I think it was poorly posed as your energy source is poorly choosen. I think with my choice of energy source the answer becomes "as many as you'd like". Up to the load limit of the reactor. Am currently looking for a datasheet, will post back when I find it.

placeholder
  • 29,982
  • 10
  • 63
  • 110
  • 2
    Is that "alternative energy source" RoHS compiliant? – Kamil Jun 10 '14 at 21:04
  • @Kamil If you have cheap fusion, you could just drop your lead-containing electronics in it and it would fuse the lead into something no-longer-lead. Of course, it would lose energy in the process, but I somehow think that's not a limiting factor... – Stephen Collings Oct 15 '15 at 14:35
2

Energy in the storage device is \$\frac{1}{2}CV^2\$, or \$4.7 \times 10^{27}\$ Joules. Energy consumed per activation is \$1.21 \times 10^{9}\$ Joules. So you can get ~\$3.89 \times 10^{18}\$ activations, or nine cycles every second. Every second. All of them.

At first I thought that to get that much energy, you'd have to consume the universe. Then I thought the destruction might in fact be very localized, limited to merely our own galaxy. But that squared term in \$E=Mc^2\$ really dominates! We're only talking about annihilating 52 gigagrams, or the mass of the Titanic. All the energy stored in your capacitor, giving that impressively large number of activations, is comparable to the power output of the sun in a tenth of a second.

So I don't know why everyone's saying this is unreasonable!

Stephen Collings
  • 17,373
  • 17
  • 92
  • 180